Simplify ((x^2-16)/(2x-18))÷((x^2-2x-8)/(x+2))

$\frac{x^{2} - 16}{2 x - 18} \div \frac{x^{2} - 2 x - 8}{x + 2}$

To divide by a fraction, multiply by its reciprocal.

$\frac{x^{2} - 16}{2 x - 18} \cdot \frac{x + 2}{x^{2} - 2 x - 8}$

Simplify the numerator.

Rewrite $16$ as $4^{2}$ .

$\frac{x^{2} - 4^{2}}{2 x - 18} \cdot \frac{x + 2}{x^{2} - 2 x - 8}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 4$ .

$\frac{(x + 4) (x - 4)}{2 x - 18} \cdot \frac{x + 2}{x^{2} - 2 x - 8}$

Factor $2$ out of $2 x - 18$ .

Factor $2$ out of $2 x$ .

$\frac{(x + 4) (x - 4)}{2 (x) - 18} \cdot \frac{x + 2}{x^{2} - 2 x - 8}$

Factor $2$ out of $- 18$ .

$\frac{(x + 4) (x - 4)}{2 x + 2 \cdot - 9} \cdot \frac{x + 2}{x^{2} - 2 x - 8}$

Factor $2$ out of $2 x + 2 \cdot - 9$ .

$\frac{(x + 4) (x - 4)}{2 (x - 9)} \cdot \frac{x + 2}{x^{2} - 2 x - 8}$

Factor $x^{2} - 2 x - 8$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 8$ and whose sum is $- 2$ .

$- 4, 2$

Write the factored form using these integers.

$\frac{(x + 4) (x - 4)}{2 (x - 9)} \cdot \frac{x + 2}{(x - 4) (x + 2)}$

Simplify terms.

Cancel the common factor of $x - 4$ .

Factor $x - 4$ out of $(x + 4) (x - 4)$ .

$\frac{(x - 4) (x + 4)}{2 (x - 9)} \cdot \frac{x + 2}{(x - 4) (x + 2)}$

Cancel the common factor.

$\frac{(x - 4) (x + 4)}{2 (x - 9)} \cdot \frac{x + 2}{(x - 4) (x + 2)}$

Rewrite the expression.

$\frac{x + 4}{2 (x - 9)} \cdot \frac{x + 2}{x + 2}$

Multiply $\frac{x + 4}{2 (x - 9)}$ by $\frac{x + 2}{x + 2}$ .

$\frac{(x + 4) (x + 2)}{2 (x - 9) (x + 2)}$

Cancel the common factor of $x + 2$ .

Cancel the common factor.

$\frac{(x + 4) (x + 2)}{2 (x - 9) (x + 2)}$

Rewrite the expression.

$\frac{x + 4}{2 (x - 9)}$

Simplify ((x^2-16)/(2x-18))÷((x^2-2x-8)/(x+2))

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